Theorem
23
[21D] If \((a_ n)_ nβ{\mathbb {R}}\) has positive terms and is monotonic (weakly) decreasing, the series converges if and only if the series
\[ β_{n=1}^β 2^ n a_{2^ n} \]
converges.
Proof
Since the sequence \((a_ n)_ n\) is decreasing, then for \(hβ{\mathbb {N}}\)
\begin{equation} 2^{h}a_{2^{(h+1)}}β€ β_{k=2^ h+1}^{2^{(h+1)}}a_ kβ€ 2^{h}a_{2^{h}}\quad .\label{eq:32rn2lp} \end{equation}
24
We note now that
\[ β_{h=0}^ Nβ_{k=2^ h+1}^{2^{(h+1)}}a_ k = β_{n=2}^{2^{N+1}}a_ n \]
and therefore
\[ β_{h=0}^ββ_{k=2^ h+1}^{2^{(h+1)}}a_ k= \lim _{Nββ} β_{h=0}^ Nβ_{k=2^ h+1}^{2^{(h+1)}}a_ k= \lim _{Nββ} β_{n=2}^{2^{(N+1)}}a_ n {=} β_{n=2}^β a_ n \quad . \]
so we can add the terms in 24 to get
\[ β_{h=0}^β 2^{h}a_{2^{(h+1)}}β€ β_{n=2}^β a_ nβ€β_{h=0}^β 2^{h}a_{2^{h}} \]
where the term on the right is finite if and only if the one on the left is finite, because
\[ β_{h=0}^β 2^{h}a_{2^{h}}=a_ 1 + 2 β_{h=0}^β 2^{h}a_{2^{(h+1)}}\quad : \]
the proof ends by the comparison theorem